first show you only need to consider squares of functions as
f.g = 1/4 [(f+g)sqr - (f-g)sqr].
show then that you only need to consider only positive valued functions becuase f(x).g(x)=|f(x)|sqr.
then , if 0 <=f(x) <= M on [a,b] show that f sqr(x) - f sqr(y) <= 2M (f(x)-f(y)).

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–
Yemon ChoiSep 29 '10 at 7:58

1

This question has been reasked on MSE, so could be closed here without loss.
–
Mariano Suárez-Alvarez♦Sep 29 '10 at 8:07

thanks for your advice, but is there a simpler approach because i am only a second year student and we have not covered Lebesgue's characterization of Riemann integrable functions at all.
–
samSep 29 '10 at 7:53

If $f$ and $g$ are Riemann integrable over the interval $[a,b]$ then there is an $M$ such that $|f|$ and $|g|$ are both $\le M$ on $[a,b]$. The Riemann integrability of $f g$ then immediately follows from the inequality
$$|f(x)g(x)-f(x')g(x')|\le |f(x)-f(x')||g(x)|+|f(x')||g(x)-g(x')|$$ $$\le
M(|f(x)-f(x')| +|g(x)-g(x')|) $$
for all $x, x'\in [a,b]$.